Answer:
I belive it would be ture
Explanation:
It's been a while since I learned this but I think that is right.
The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration <em>a</em> is such that
4 m/s = <em>a</em> (2.5 s) → <em>a</em> = (4 m/s) / (2.5 s) = 1.6 m/s²
Then the force applied to the box has a magnitude <em>F</em> such that
<em>F</em> = (10 kg) (1.6 m/s²) = 16 N
Answer:
the rate of turn at any airspeed is dependent upon the horizontal lift component
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m
Formula:

Solving:



Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m
Formula:

Solving:(Energy associated with this stretching)




Answer:
Moving a unit "positive" test charge from A to B will result in a reduction in potential
V = K Q / R potential at a point
V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q
V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8
V2 - V1 = -4.17 * 562.5 J/C
V = - 2346 Volts